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Retired Moderator D
Joined: 25 Feb 2013
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Mr. Bond has forgotten his n digit locker code  [#permalink]

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Question Stats: 39% (03:11) correct 61% (03:13) wrong based on 92 sessions

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Mr. Bond has forgotten his $$n$$ digit locker code. He remembers that the first three digits are either $$853$$ or $$847$$ and there was a number that appeared only once in the code. If Mr. Bond were to use a trial and error process to open his locker, what is the minimum number of trials he has to make before he can be certain to succeed?

A.) Mr. Bond recalled that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than $$10$$. He also recalled that the digit that appeared only once was square of a single digit odd prime number and the locker code was not a multiple of two.

B.) The locker code is odd, and the number nine appears once in the code. $$n<14$$ and $$n=6k+1$$, where $$k$$ is any integer.

$$Note:$$ Kudos for all Correct EXPLANATIONS Intern  B
Joined: 31 Oct 2016
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Re: Mr. Bond has forgotten his n digit locker code  [#permalink]

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A clearly tells that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than 10. Only case 2*7 because 2*3, 2*5 are both =/< 10 so only left with one possibility.

B on the other hand tells us that n<14 and n=6k+1. This gives us two possibilities: 7 and 13.

The question is asking to find the minimum number of trials before he can be certain to succeed and if I don't get into the details of other calculations. this is good enough to rule out B and keep A.

IMO - A
Retired Moderator D
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Mr. Bond has forgotten his n digit locker code  [#permalink]

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Gatt33 wrote:
A clearly tells that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than 10. Only case 2*7 because 2*3, 2*5 are both =/< 10 so only left with one possibility.

B on the other hand tells us that n<14 and n=6k+1. This gives us two possibilities: 7 and 13.

The question is asking to find the minimum number of trials before he can be certain to succeed and if I don't get into the details of other calculations. this is good enough to rule out B and keep A.

IMO - A

Hi Gatt33
Can you provide a more detailed explanation for Option A as to how we can use Statement A to get the minimum value or why not Option E because in the GMAT we will not have the luxury to know the OA Current Student S
Joined: 22 Apr 2017
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GMAT 1: 620 Q46 V30 GMAT 2: 620 Q47 V29 GMAT 3: 630 Q49 V26 GMAT 4: 690 Q48 V35 GPA: 3.7
Mr. Bond has forgotten his n digit locker code  [#permalink]

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niks18 wrote:
Gatt33 wrote:
A clearly tells that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than 10. Only case 2*7 because 2*3, 2*5 are both =/< 10 so only left with one possibility.

B on the other hand tells us that n<14 and n=6k+1. This gives us two possibilities: 7 and 13.

The question is asking to find the minimum number of trials before he can be certain to succeed and if I don't get into the details of other calculations. this is good enough to rule out B and keep A.

IMO - A

Hi Gatt33
Can you provide a more detailed explanation for Option A as to how we can use Statement A to get the minimum value or why not Option E because in the GMAT we will not have the luxury to know the OA Hi niks18,
IMO, from A, we get that the length of code is 7 digits & digit 9 is not repeated in the code. So we can find out the total number of trials possible. As far as minimum number of trials is concerned, it could be anything between 1 & total combination possible.

HiBunuel, could you kindly shed some light on it.

Regards
Retired Moderator D
Joined: 25 Feb 2013
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Re: Mr. Bond has forgotten his n digit locker code  [#permalink]

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1
niks18 wrote:
Mr. Bond has forgotten his $$n$$ digit locker code. He remembers that the first three digits are either $$853$$ or $$847$$ and there was a number that appeared only once in the code. If Mr. Bond were to use a trial and error process to open his locker, what is the minimum number of trials he has to make before he can be certain to succeed?

OE

To solve this Permutation problem we need three values-
a) the value of n = no of digit in the locker code
b) the digit that appeared only once
c) whether the locker code was ODD or EVEN because permutation will get impacted based on even/odd combination as we have one number that can be used only once

Statement 1:
Quote:
Mr. Bond recalled that the number of digits in the code was a single digit prime number which, when multiplied by another prime number, returned an even number greater than $$10$$

this implies that $$n=7$$ because $$7*2=14>10$$

Quote:
He also recalled that the digit that appeared only once was square of a single digit odd prime number

this implies that the digit that appeared only once was $$9$$=$$3^2$$

Quote:
and the locker code was not a multiple of two.

this implies that the code is ODD

Thus we got all the information we were looking for. Hence Sufficient. Note: this is a DS problem hence we don't need an exact answer here.

Statement 2:
Quote:
The locker code is odd, and the number nine appears once in the code. $$n<14$$ and $$n=6k+1$$, where $$k$$ is any integer.

From this we get the information that the code is ODD and 9 appears only once, but the value of $$n$$ cannot be determined,
if $$k=1$$, $$n=7$$ & if $$k=2$$, $$n=13$$. Thus two values of n are possible. Hence Insufficient

Option A
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FYI - calculation for number of trials

There are two possible cases. The number $$9$$ comes at the end, or it comes at position 4th, 5th, or 6th.

For the first case, the number would look like: $$853 - - - 9$$ or $$847 - - - 9$$

In both these cases, the blanks can be occupied by any of the available $$9$$ digits (0, 1, 2, ..., 8).

Thus, total possible numbers would be $$2 × (9 × 9 × 9) = 1458$$.

For the second case, the number $$9$$ can occupy any of the given position 4th, 5th, or 6th, and there shall be an odd number at position 7th.

Thus, the total number of ways shall be $$2[3(9 × 9 × 4)] = 1944$$. Hence, answer is $$1458+1944=3402$$.
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Re: Mr. Bond has forgotten his n digit locker code  [#permalink]

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_________________ Re: Mr. Bond has forgotten his n digit locker code   [#permalink] 25 Feb 2019, 01:05
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